Euler's sum of powers conjecture#Counterexamples

Euler was aware of the equality {{nowrap|59{{sup|4}} + 158{{sup|4}} {{=}} 133{{sup|4}} + 134{{sup|4}}}} involving sums of four fourth powers; this, however, is not a counterexample because no term is isolated on one side of the equation. He also provided a complete solution to the four cubes problem as in Plato's number {{nowrap|3{{sup|3}} + 4{{sup|3}} + 5{{sup|3}} {{=}} 6{{sup|3}}}} or the taxicab number 1729.{{cite book |editor-last=Dunham |editor-first=William |year=2007 |title=The Genius of Euler: Reflections on His Life and Work |publisher=The MAA |isbn=978-0-88385-558-4 |page=220 |url=https://books.google.com/books?id=M4-zUnrSxNoC&pg=PA220}}{{cite web |last=Titus, III |first=Piezas |year=2005 |title=Euler's Extended Conjecture |url=http://www.oocities.org/titus_piezas/Equalsums.htm }} The general solution of the equation $x\_1^3+x\_2^3=x\_3^3+x\_4^3$

is

$$\backslash begin\{align\}$$

x_1 &=\lambda( 1-(a-3b)(a^2+3b^2)) \\[2pt]

x_2 &=\lambda( (a+3b)(a^2+3b^2)-1 )\\[2pt]

x_3 &=\lambda( (a+3b)-(a^2+3b^2)^2 )\\[2pt]

x_4 &= \lambda( (a^2+3b^2)^2-(a-3b))

\end{align}

where {{mvar|a}}, {{mvar|b}} and $\{\backslash lambda\}$ are any rational numbers.

Euler's conjecture was disproven by L. J. Lander and T. R. Parkin in 1966 when, through a direct computer search on a CDC 6600, they found a counterexample for {{math|k {{=}} 5}}.{{cite journal |last1=Lander |first1=L. J. |last2=Parkin |first2=T. R. |year=1966 |title=Counterexample to Euler's conjecture on sums of like powers |journal=Bull. Amer. Math. Soc. |doi=10.1090/S0002-9904-1966-11654-3 |volume=72 |issue=6 |page=1079|doi-access=free }} This was published in a paper comprising just two sentences. A total of three primitive (that is, in which the summands do not all have a common factor) counterexamples are known:

$$\backslash begin\{align\}$$

144^5 &= 27^5 + 84^5 + 110^5 + 133^5 \\

14132^5 &= (-220)^5 + 5027^5 + 6237^5 + 14068^5 \\

85359^5 &= 55^5 + 3183^5 + 28969^5 + 85282^5

\end{align}

(Lander & Parkin, 1966); (Scher & Seidl, 1996); (Frye, 2004).

In 1988, Noam Elkies published a method to construct an infinite sequence of counterexamples for the {{math|k {{=}} 4}} case.{{cite journal |last=Elkies |first=Noam |authorlink=Noam Elkies |year=1988 |title=On A⁴ + B⁴ + C⁴ = D⁴ |journal=Mathematics of Computation |doi=10.1090/S0025-5718-1988-0930224-9 |mr=0930224 |jstor=2008781 |volume=51 |issue=184 |pages=825–835 |url=https://www.ams.org/journals/mcom/1988-51-184/S0025-5718-1988-0930224-9/S0025-5718-1988-0930224-9.pdf |doi-access=free }} His smallest counterexample was

$$20615673^4\; =\; 2682440^4\; +\; 15365639^4\; +\; 18796760^4.$$

A particular case of Elkies' solutions can be reduced to the identity{{cite web |title=Elkies' a⁴+b⁴+c⁴ = d⁴ |url=https://groups.google.com/group/sci.math/browse_thread/thread/15beef75eaddcb1b?hl=en#}}{{cite book|year = 2010|chapter = Sums of Three Fourth Powers (Part 1)|title = A Collection of Algebraic Identities|first = Tito|last = Piezas III|chapter-url = http://sites.google.com/site/tpiezas/014|access-date = April 11, 2022}}

$$(85v^2\; +\; 484v\; -\; 313)^4\; +\; (68v^2\; -\; 586v\; +\; 10)^4\; +\; (2u)^4\; =\; (357v^2\; -\; 204v\; +\; 363)^4,$$

where

$$u^2\; =\; 22030\; +\; 28849v\; -\; 56158v^2\; +\; 36941v^3\; -\; 31790v^4.$$

This is an elliptic curve with a rational point at {{math|v₁ {{=}} −{{sfrac|31|467}}}}. From this initial rational point, one can compute an infinite collection of others. Substituting {{math|v₁}} into the identity and removing common factors gives the numerical example cited above.

In 1988, Roger Frye found the smallest possible counterexample

$$95800^4\; +\; 217519^4\; +\; 414560^4\; =\; 422481^4$$

for {{math|k {{=}} 4}} by a direct computer search using techniques suggested by Elkies. This solution is the only one with values of the variables below 1,000,000.{{citation

| last = Frye | first = Roger E.

| year = 1988

| title = Proceedings of Supercomputing 88, Vol.II: Science and Applications

| contribution = Finding 95800⁴ + 217519⁴ + 414560⁴ = 422481⁴ on the Connection Machine

| doi = 10.1109/SUPERC.1988.74138

| pages = 106–116| s2cid = 58501120

}}

File:Plato_number.svg

{{main article|Lander, Parkin, and Selfridge conjecture}}

In 1967, L. J. Lander, T. R. Parkin, and John Selfridge conjectured{{cite journal |last1=Lander |first1=L. J. |last2=Parkin |first2=T. R. |last3=Selfridge |first3=J. L. |year=1967 |title=A Survey of Equal Sums of Like Powers |journal=Mathematics of Computation |doi=10.1090/S0025-5718-1967-0222008-0 |jstor=2003249 |volume=21 |issue=99 |pages=446–459 |doi-access=free }} that if

:$\backslash sum\_\{i=1\}^\{n\}\; a\_i^k\; =\; \backslash sum\_\{j=1\}^\{m\}\; b\_j^k$,

where {{math|a_i ≠ b_j}} are positive integers for all {{math|1 ≤ i ≤ n}} and {{math|1 ≤ j ≤ m}}, then {{math|m + n ≥ k}}. In the special case {{math|m {{=}} 1}}, the conjecture states that if

:$\backslash sum\_\{i=1\}^\{n\}\; a\_i^k\; =\; b^k$

(under the conditions given above) then {{math|n ≥ k − 1}}.

The special case may be described as the problem of giving a partition of a perfect power into few like powers. For {{math|k {{=}} 4, 5, 7, 8}} and {{math|n {{=}} k}} or {{math|k − 1}}, there are many known solutions. Some of these are listed below.

See {{OEIS2C|A347773}} for more data.

$$3^3\; +\; 4^3\; +\; 5^3\; =\; 6^3$$ (Plato's number 216)

This is the case {{math|1=a = 1}}, {{math|1=b = 0}} of Srinivasa Ramanujan's formula{{cite web| url = http://mathworld.wolfram.com/DiophantineEquation3rdPowers.html| title = MathWorld : Diophantine Equation--3rd Powers}}

$$(3a^2+5ab-5b^2)^3\; +\; (4a^2-4ab+6b^2)^3\; +\; (5a^2-5ab-3b^2)^3\; =\; (6a^2-4ab+4b^2)^3$$

A cube as the sum of three cubes can also be parameterized in one of two ways:

$$\backslash begin\{align\}$$

a^3(a^3+b^3)^3 &= b^3(a^3+b^3)^3+a^3(a^3-2b^3)^3+b^3(2a^3-b^3)^3 \\[6pt]

a^3(a^3+2b^3)^3 &= a^3(a^3-b^3)^3+b^3(a^3-b^3)^3+b^3(2a^3+b^3)^3.

\end{align}

The number 2,100,000³ can be expressed as the sum of three positive cubes in nine different ways.

$$\backslash begin\{align\}$$

422481^4 &= 95800^4 + 217519^4 + 414560^4 \\[4pt]

353^4 &= 30^4 + 120^4 + 272^4 + 315^4

\end{align}

(R. Frye, 1988); (R. Norrie, smallest, 1911).

$$\backslash begin\{align\}$$

144^5 &= 27^5 + 84^5 + 110^5 + 133^5 \\[2pt]

72^5 &= 19^5 + 43^5 + 46^5 + 47^5 + 67^5 \\[2pt]

94^5 &= 21^5 + 23^5 + 37^5 + 79^5 + 84^5 \\[2pt]

107^5 &= 7^5 + 43^5 + 57^5 + 80^5 + 100^5

\end{align}

(Lander & Parkin, 1966);{{cite web |url=https://www.youtube.com/watch?v=AO-W5aEJ3Wg | archive-url=https://ghostarchive.org/varchive/youtube/20211211/AO-W5aEJ3Wg| archive-date=2021-12-11 | url-status=live|title=Euler's and Fermat's last theorems, the Simpsons and CDC6600 |author=Burkard Polster |website=YouTube |date=March 24, 2018 |author-link=Burkard Polster |type=video |access-date=2018-03-24}}{{cbignore}}{{cite web| url = https://mathworld.wolfram.com/DiophantineEquation5thPowers.html| title = MathWorld: Diophantine Equation--5th Powers}}{{cite web| url = https://pat7.com/jp/s515-10007-t| title = A Table of Fifth Powers equal to Sums of Five Fifth Powers}} (Lander, Parkin, Selfridge, smallest, 1967); (Lander, Parkin, Selfridge, second smallest, 1967); (Sastry, 1934, third smallest).

It has been known since 2002 that there are no solutions for {{math|1=k = 6}} whose final term is ≤ 730000.Giovanni Resta and Jean-Charles Meyrignac (2002). [https://www.ams.org/journals/mcom/2003-72-242/S0025-5718-02-01445-X/S0025-5718-02-01445-X.pdf The Smallest Solutions to the Diophantine Equation $a^6+b^6+c^6+d^6+e^6=x^6+y^6$], Mathematics of Computation, v. 72, p. 1054 (See further work section).

$$568^7\; =\; 127^7\; +\; 258^7\; +\; 266^7\; +\; 413^7\; +\; 430^7\; +\; 439^7\; +\; 525^7$$

(M. Dodrill, 1999).{{cite web| url = https://mathworld.wolfram.com/DiophantineEquation7thPowers.html| title = MathWorld: Diophantine Equation--7th Powers}}

$$1409^8\; =\; 90^8\; +\; 223^8\; +\; 478^8\; +\; 524^8\; +\; 748^8\; +\; 1088^8\; +\; 1190^8\; +\; 1324^8$$

(S. Chase, 2000).{{cite web| url = https://mathworld.wolfram.com/DiophantineEquation8thPowers.html| title = MathWorld: Diophantine Equation--8th Powers}}

• Jacobi–Madden equation
• Prouhet–Tarry–Escott problem
• Beal conjecture
• Pythagorean quadruple
• Generalized taxicab number
• Sums of powers, a list of related conjectures and theorems

{{reflist}}

• Tito Piezas III, [http://sites.google.com/site/tpiezas/Home/ A Collection of Algebraic Identities] {{Webarchive|url=https://web.archive.org/web/20111001021837/http://sites.google.com/site/tpiezas/Home |date=2011-10-01 }}
• Jaroslaw Wroblewski, [http://www.math.uni.wroc.pl/~jwr/eslp/ Equal Sums of Like Powers]
• Ed Pegg Jr., [https://web.archive.org/web/20080410224256/http://www.maa.org/editorial/mathgames/mathgames_11_13_06.html Math Games, Power Sums]
• James Waldby, [http://pat7.com/jp/s515-10007-t A Table of Fifth Powers equal to a Fifth Power (2009)]
• R. Gerbicz, J.-C. Meyrignac, U. Beckert, [https://arxiv.org/abs/1108.0462 All solutions of the Diophantine equation a⁶ + b⁶ = c⁶ + d⁶ + e⁶ + f⁶ + g⁶ for a,b,c,d,e,f,g < 250000 found with a distributed Boinc project]
• [http://euler.free.fr/ EulerNet: Computing Minimal Equal Sums Of Like Powers]
• {{MathWorld |title=Euler's Sum of Powers Conjecture |urlname=EulersSumofPowersConjecture}}
• {{MathWorld |title=Euler Quartic Conjecture |urlname=EulerQuarticConjecture}}
• {{MathWorld |title=Diophantine Equation--4th Powers |urlname=DiophantineEquation4thPowers}}
• [https://web.archive.org/web/20071105172444/http://library.thinkquest.org/28049/Euler%27s%20conjecture.html Euler's Conjecture] at library.thinkquest.org
• [http://www.mathsisgoodforyou.com/conjecturestheorems/eulerconjecture.htm A simple explanation of Euler's Conjecture] at Maths Is Good For You!

{{Leonhard Euler}}

Category:Diophantine equations

Category:Disproved conjectures

Category:Leonhard Euler